A deck of forty cards consists of four 1's, four 2's,..., and four 10's. A matching pair (two cards with the same number) is removed from the deck. Given that these cards are not returned to the deck, let $m/n$ be the probability that two randomly selected cards also form a pair, where $m$ and $n$ are relatively prime positive integers. Find m + n.
My logic was that well given you take away 2 matching cards you have a net of 38 cards. The odds that you take away from the same number pile is 2/38 * 1/37. We must also add the odds that you take away from different numbers (remaining 9 are left) which makes it 9 *(4/38 * 3/37). This yields 55/703 (which is the correct answer). M +. n = 758
However, I figured we should multiply the entire fraction by 10 since a pair from any of the 10 numbers can be taken out first, which is wrong. However, I cant understand why.
I think I have a solution that might shed some light on why we wouldn't multiply by $10$. Let us consider the simpler case of just drawing a pair from the deck without removing a pair. We would get $$10\times\left(\dfrac{4}{40}+\dfrac{3}{39}\right)$$ The $\frac{4}{40}$ is the probability of drawing the first card from the deck and then $\frac{3}{39}$ is for drawing the matching pair. The $10$ our front is the checking for each of the $10$ numbers. I believe that this is the $10$ that you are picturing in this problem that you would like to account for.
Let us now consider this case of removing a pair. I will rewrite the answer you gave $$1\times\left(\dfrac{2}{38}+\dfrac{1}{37}\right)+9\times\left(\dfrac{4}{38}+\dfrac{3}{37}\right)$$ Here we have simply take the $10$ and split it up into the the case of the $1$ pair removed and the $9$ left in. So the $10$ that you are concerned about is already accounted for in the calculation.